Using \(K^2=\mathbb {I}\) , one can easily check that
these operators satisfy \(P^2=P\) , \(\tilde{P}^2=\tilde{P}\)  and \(P\tilde{P}=0=\tilde{P}P\) .
The Nijenhuis tensor (REF ) for \(A\equiv K\)  can be rewritten in terms of these projectors as
[1]}
\(N_K(X,Y)\equiv N_{P}(X,Y)+N_{\tilde{P}}(X,Y),\) 
